Question: Is ${351462}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {351462}= &&{3}\cdot100000+ \\&&{5}\cdot10000+ \\&&{1}\cdot1000+ \\&&{4}\cdot100+ \\&&{6}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {351462}= &&{3}(99999+1)+ \\&&{5}(9999+1)+ \\&&{1}(999+1)+ \\&&{4}(99+1)+ \\&&{6}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {351462}= &&\gray{3\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {3}+{5}+{1}+{4}+{6}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${351462}$ is divisible by $3$ if ${ 3}+{5}+{1}+{4}+{6}+{2}$ is divisible by $3$ Add the digits of ${351462}$ $ {3}+{5}+{1}+{4}+{6}+{2} = {21} $ If ${21}$ is divisible by $3$ , then ${351462}$ must also be divisible by $3$ ${21}$ is divisible by $3$, therefore ${351462}$ must also be divisible by $3$.